The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 × 1 0 6 ( A ) 2 ( m ) 2 . The current through the outer portion of the wire between radial distances 2 r and r is x π A ; where

Question

The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 × 1 0 6 ( A ) 2 ( m ) 2 . The current through the outer portion of the wire between radial distances 2 r ​ and r is x π A ; where x is

TestHub · Accepted Answer

Current = J × Area = J × π [ R 2 − ( 2 R ​ ) ] 2 = J π 4 3 R 2 ​ = 1 0 6 × π × 3 × ( 4 4 × 1 0 − 3 ​ ) 2 ⇒ π × 3 × 4 16 ​ = 12 π A Hence, the value of x = 12 .

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